Calculate the molar concentration of the cl⁻ ions in 0.87 m mgcl2(aq).

The compound MgCl2(aq) is ionic which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MgCl2 that dissolves.
MgCl2(s) --> Mg+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.87 mol MgCl2/1L × 2 mol Cl⁻ / 1 mol MgCl2 = 1.7 M
The answer to this question is [Cl⁻] = 1.7 M

Dissociation then hydration dissolving